find the equation of contact plane with the chart: z = ln(1+xy) in the point (1,2,ln3). You also
have to find the equation and parametrization normal line on the chart in given point.
Solution
F(x,y,z) = z - ln(1+xy)
dF/dx = -y/(1+xy) = -2/(1+2) = -2/3
dF/dy = -x/(1+xy) = -1/(1+2) = -1/3
dF/dz = 1
Therefore the contact plane equation is:
-2/3 (x - 1) - 1/3 (y - 2) + 1(z - ln(3)) = 0
or:
2x + y - 3z = 2 + 2 - 3ln(3)
2x + y - 3z = 4 - ln(3)